The work functions of cesium (Cs) and lithium (Li) metals are 1.9 eV and 2.5 eV, respectively. If we incident a light of wavelength 550 nm on these two metal surface, then photo-electric effect is possible for the case of

To determine if the photoelectric effect occurs for cesium (Cs) and lithium (Li) under incident light of wavelength 550 nm, we compare the energy of incident photons with the work function ($\phi$) of each metal.

Step 1: Calculate the Energy of the Incident Photon

The energy of a photon is given by the equation:

$$E = \frac{hc}{\lambda}$$

where:

$h = 6.626 \times 10^{-34}$ J·s (Planck’s constant),

$c = 3.0 \times 10^8$ m/s (speed of light),

$\lambda = 550$ nm $= 550 \times 10^{-9}$ m.

First, calculate the photon energy in joules:

$$E = \frac{(6.626 \times 10^{-34}) (3.0 \times 10^8)}{550 \times 10^{-9}}$$

 $$E = \frac{1.9878 \times 10^{-25}}{550 \times 10^{-9}}$$ 

$$E = 3.615 \times 10^{-19} \text{ J}$$

Convert this to electron volts (eV) using $1 \text{ eV} = 1.6 \times 10^{-19}$ J:

$$E = \frac{3.615 \times 10^{-19}}{1.6 \times 10^{-19}}$$

 $$E \approx 2.26 \text{ eV}$$ 

Step 2: Compare Photon Energy with Work Functions

Cesium ($\phi_{\text{Cs}} = 1.9$ eV)
Since $E_{\text{photon}} = 2.26$ eV is greater than $\phi_{\text{Cs}} = 1.9$ eV,photoelectric emission occurs.

Lithium ($\phi_{\text{Li}} = 2.5$ eV)
Since $E_{\text{photon}} = 2.26$ eV is less than $\phi_{\text{Li}} = 2.5$ eV, photoelectric emission does not occur.

Conclusion:

Photoelectric effect is possible only for Cesium (Cs), but not for Lithium (Li).